factor 4.9t^2-10t+2.5

Question

anonymous · Answer

using quadratic formula?

anonymous · Answer

yes

mathmale · Answer

Certainly you can factor using the quadratic formula, but I don't see that formula as being necessary.  Note that 0.49 is the square of 0.7.

anonymous · Answer

$$10\pm \sqrt{10}/18$$

anonymous · Answer

Well, 
Given the general form 
$$a{x}^{2}+bx+c=0 $$

there exists two solutions where:
$$x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}$$

In this case, a=4.9, b=−10 and c=2.5

$$x=\frac{-(-10) \pm\sqrt{{(-10)}^{2}-4(4.9)(2.5)}}{2(4.9)}$$

anonymous · Answer

it would be $$t=\frac{10+\sqrt{{10}^{2}-49}}{9.8},\frac{10-\sqrt{{10}^{2}-49}}{9.8}$$

anonymous · Answer

ok. so t= $$10\pm \sqrt{51}\div9.8$$

mathmale · Answer

Close.  You must (absolutely must) enclose that 10 plus or minus Sqrt(51) inside parentheses.

How would you check your answers?

anonymous · Answer

would i plug that into the equation for x?

anonymous · Answer

yes, because there are two solutions.